I am new to DFT, especially in doing DFT for magnetic materials. I recently came across this paper which indicated the calculation of the J value in the case of antiferromagentic materials as per the formula.

$$E_{HS}- E_{BS} = -J(2S_{1}S_{2}-S_{2}) $$

The paper describes the terms as follows:

$E_{HS}$ : Energy of the triple state {High Spin}: I have calculated this by constrained magnetization method in Quantum ESPRESSO.

$E_{BS}$ : Energy of the Broken symmetry state {Low spin}: I have doubts regarding what this is; I do believe its the anti-ferromagnetic state {total magnetization = zero}. Correct me if Im wrong!

The $S_{1}$ and $S_{2}$ in the paper : where $S_1$ and $S_2$ are the total spins of the paramagnetic centers and $S_1$>$S_2$ has been assumed for heterodinuclear complexes using the Heisenberg Hamiltonian. I didnt quite understand this : How can I calculate this?

Thank you in Advance!!


1 Answer 1


In the attached paper they have used the broken symmetry method to calculate the exchange.

Assuming that you're dealing with a 1D chain of spins. you can do this using quantum ESPRESSO via the following steps:

  1. Create a 2x1x1 super cell from your CIF file using VESTA (assuming that your 1d chain is along the first basis vector)

  2. For the SCF calculation using QE, you need to consider two cases a ferromagnetic state and an antiferromagnetic state. Here the magnetic moment on the atom (one in each unit cell) should be constrained as follows in the &SYSTEM name card:

For the AFM case:

  starting_magnetization(1) = 1.0
  starting_magnetization(2) = -1.0
  constrained_magnetization = 'atomic'

For the FM case:

  starting_magnetization(1) = 1.0
  starting_magnetization(2) = 1.0
  constrained_magnetization = 'atomic'
  1. The difference in energy should give you the J value:

    $$ J = E_{FM}-E_{AFM}\tag{1}$$

I hope this is correct!

  • $\begingroup$ +1 the one minor edit I made was only so that if anyone needs to refer to that equation they can say "as seen in Eq. 1 of this answer" rather than "as seen in the equation in this answer". $\endgroup$ May 16, 2021 at 21:58

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